A stunt flier is moving at 15m/s parallel to the flat ground 100m below,how large must the distance x from plane to target be if a sack of flour released from the place is to strike the target?
find the fall time (t) ... 1/2 g t^2 = 100
the distance is ... 15 t
10
To determine the distance x from the plane to the target, we can use the equation of motion. The vertical motion of the sack of flour follows the equation:
y = y0 + v0yt - (1/2)gt^2
Where:
y = final vertical position (distance to the target) [unknown]
y0 = initial vertical position (height of the plane) = 100m
v0y = vertical component of the initial velocity = 0m/s (since the plane is moving parallel to the ground)
g = acceleration due to gravity = 9.8m/s^2
t = time it takes for the sack of flour to reach the target [unknown]
To find the time (t) for the sack of flour to reach the target, we need to use the horizontal motion of the plane. Since the plane is moving parallel to the ground, there is no horizontal acceleration, and the horizontal distance covered by the sack of flour (x) is equal to the horizontal component of the velocity (v0x) multiplied by time (t).
Given that the plane's velocity is 15m/s, the horizontal component of velocity (v0x) is 15m/s.
Now, we can use the horizontal motion equation:
x = v0xt
Rearranging the equation to solve for time:
t = x / v0x
Substitute this value of time (t) into the vertical motion equation:
y = y0 - (1/2)gt^2
Substitute the given values:
y = 100m
g = 9.8m/s^2
Now we can solve for x by substituting the expression for time (t) into the vertical motion equation:
y = y0 - (1/2)g(x / v0x)^2
Simplify the equation:
100 = 100 - (1/2)(9.8)(x / 15)^2
Simplify further:
1 = (1/2)(9.8)(x / 15)^2
Rearrange and solve for x:
[(9.8)(x / 15)^2] / 2 = 1
Simplify:
(9.8)(x / 15)^2 = 2
Solve for (x / 15)^2:
(x / 15)^2 = 2 / 9.8
Take the square root of both sides:
x / 15 = √(2 / 9.8)
Multiply both sides by 15:
x = 15 * √(2 / 9.8)
Calculate the value:
x ≈ 6.09m
Therefore, the distance x from the plane to the target must be approximately 6.09 meters in order for the sack of flour released from the plane to strike the target.
To determine the distance x from the plane to the target, we need to consider the horizontal motion and vertical motion separately.
Let's first look at the horizontal motion. The stunt flier is moving parallel to the ground, so the horizontal velocity of the sack of flour will match that of the plane, which is 15 m/s.
Now, let's consider the vertical motion. The sack of flour will fall under the influence of gravity, just like any other object in free fall. We can use the equations of motion to find the time it takes for the sack of flour to hit the ground.
The vertical displacement of the sack of flour is the height from the plane to the target, which is 100 meters. We can use the equation of motion for vertical displacement:
y = vt + (1/2)gt²
Here, y represents the vertical displacement, v is the initial vertical velocity (which is zero since the sack is dropped), t is the time, g is the acceleration due to gravity (approximately 9.8 m/s²), and we need to solve for t.
Substituting the known values:
100 = 0t + (1/2)(9.8)(t²)
Simplifying the equation:
100 = 4.9t²
Rearranging the equation:
t² = 100/4.9
t ≈ √(20.41) ≈ 4.52 seconds
Now that we have the time it takes for the sack of flour to fall, we can calculate the horizontal distance using the horizontal velocity and time:
x = vt
x = 15 m/s * 4.52 s
x ≈ 67.8 meters
Therefore, the distance x from the plane to the target must be approximately 67.8 meters for the sack of flour to strike the target.